Question

A cylindrical tank, 14 m in diameter and 10 m high, is full of water. The tank is standing upright on a flat circular end. How much work is required to pump half the water over the top of the tank? (The density of water is 1000 kg/m³ ). (Use symbolic notation and fractions where needed.)

Answer 1

The amount of work required to pump half the water over the top of the tank is approximately
`$76,995 \mathrm{~J}$` when rounded to the nearest whole number.

The tank is a cylinder with a diameter of
`$14 \mathrm{~m}$`, so the radius
`$(r)$` is
`$7 \mathrm{~m}$`.

The height
`$(h)$` of the tank is
`$10 \mathrm{~m}$`.

The volume
`$(V)$` of the water in the tank is given by the formula
`$V = πr^2h$`.

Plugging in the values, we get
`$V = π(7^2)(10) = 490π \mathrm{~m}^3$`.

To find the volume of half the water, we calculate
`$V/2 = (490π)/2 = 245π \mathrm{~m}^3$`.

The density
`$(\rho)$` of water is
`$1000 \mathrm{~kg/m^3}$`.

The mass
`$(m)$` of the water is given by the formula
`$m = \rho * V$`.

Plugging in the values, we get
`$m = 1000 * 245π \mathrm{~kg}$`.

The weight
`$(W)$` of the water is given by the formula
`$W = m * g$`, where
`$g$` is the gravitational acceleration (
`$9.8 \mathrm{~m/s^2}$`).

Plugging in the values, we have
`$W = (1000 * 245π) * 9.8 \mathrm{~N}$`.

The work
`$(\mathrm{W})$` done to lift the water over the top of the tank is given by the formula
`$W = F * d$`, where
`$F$` is the force (weight) and
`$d$` is the distance (height of the tank).

Plugging in the values, we get
`$W = (1000 * 245π) * 9.8 * 10 \mathrm{~J}$`.

So, the amount of work required to pump half the water over the top of the tank is
`$(1000 * 245π) * 9.8 * 10 \mathrm{~J}$`. This can be simplified as
`$24,500π \mathrm{~J}$` or approximately
`$76,995 \mathrm{~J}$` when rounded to the nearest whole number.